Link between the hairy ball theorem and the fundamental theorem of algebra I read in the book "Concepts of modern mathematics" by Ian Stewart that it was possible to proof the fundamental theorem of algebra using the hairy ball theorem (complete reference to the page is in the following quora question : https://www.quora.com/How-can-the-hairy-ball-theorem-be-used-to-prove-the-fundamental-theorem-of-algebra). I am not aware of such a proof. It is not obvious to me (compactification does not seem to lead to a satisfactory answer) and I would be delighted if someone could help me with that.
 A: Maybe this is an idea. It suffices to prove that any polynomial $f\in \mathbf C[z]$ of even degree $2d\geq2$ has a zero in $\mathbf C$. Assume that it has no zero. Then the rational section
$$
s=f\cdot\left(\tfrac{\partial}{\partial z}\right)^{\otimes d}
$$
of the $d$-th tensor power $T^{\otimes d}$ of the holomorphic tangent bundle on the complex projective line $\mathbf P^1(\mathbf C)$ is a nowhere vanishing global section. This is so since the holomorphic vector field $\partial/\partial z$ does not vanish on $\mathbf C$, has a zero of order $2$ at $\infty$, the polynomial $f$ does not vanish on $\mathbf C$, and has a pole of order $2d$ at $\infty$. Consider the section $s$ as a topological surface in the complex line bundle $T^{\otimes d}$ over $\mathbf P^1(\mathbf C)$. As image of the sphere $S^2$, it is homeomorphic to $S^2$ and does not have any nontrivial topological coverings. Its inverse image under the power-$d$-map
$$
T\rightarrow T^{\otimes d}
$$
from the complex line bundle $T$ into the complex line bundle $T^{\otimes d}$ is, therefore, a disjoint union of $d$ disjoint copies of the sphere $S^2$ in the complement of the zero section of the tangent bundle of $S^2$. Each would define a nowhere vanishing vector field on $S^2$. Contradiction by the hairy ball theorem.
A: This is a (very) partial answer: Suppose the polynomial is real and of odd degree. Then this polynomial is the characteristic polynomial of some matrix acting in an odd number of dimensions. Restricting this matrix to the sphere and then projecting onto the tangent space of the sphere defines a vector field on an even dimensional sphere, hence by the Hairy Ball Theorem it vanishes somewhere. The place where it vanishes is an eigenvector of the matrix, hence there is a corresponding eigenvalue, hence the polynomial has a root.
I'm not sure if something similar can be done for arbitrary polynomials though.
